Hi, have you tried solving the square binomial and then apply vertex formula ( ) and then substitute y = 0. Finally you can find and
I need help with some algebra !!
Here is the problem: find the Vertex of the Parabola y= -2(x-2)^2+10
I need the Vertex. ( I got 1.99?) x= 2.0 ?
Then I have to find the horizontal intercepts of the Parabola where y = 0.
Then I have to put it in factored form of the parabola in Form y = a( x - x1 )( x - x2 ))
I came up with a Vertex of 1.99 rounded to 2.0? But have no idea where or what to do next. I tried using my graphing calculator and entered the equation into y = but it comes up error. I do not know what I am doing wrong...
How in the world did you get 1.99? If x= 2 then x- 2= 0 so -2(x- 2)^2= 0 and y= 10. For any other x, x- 2 is non-zero so (x- 2)^2 is positive and -2(x- 2)^2 is negative so y is less than 10.
Yes, the "horizontal" or "x" intercepts, where the graph crosses the x-axis, are where y= 0. So you need to solve the equation -2(x- 2)^2+ 10= 0 That is, of course, the same as
-2(x- 2)^2= -10 or, dividing by -2, (x- 2)^2= 5.
Obviously a(x- x1)(x- x2) will be 0 where x= x1 or x= x2 so those have to be the points where y= 0 you just found.
Thanks for answering but I still do not understand how to come up with the answer to my problems. Sorry I just don't get this algebra stuff. I have 3 weeks left in my college algebra class before I graduate. Math is by far my worst subject. I was hoping someone could tell me step by step what I need to do ? Cause I'm lost. I appreciate you taking the time to answer my post.
Actually there are not many steps to follow and the idea is help you to understand. As I said, have you tried using the vertex formula I wrote?
I guess you have solved this types of equations (or at least have an example to follow), so I suggest you to expand if you're completely lost. That will lead you to the traditional form of quadratic functions.
There are a few formulas that are meant to use in order to find some special points from the curve.
Let's say your function looks like: you could:
Find its vertex: and
Find its roots (x values which makes y = 0):
As you can see the symbol indicates that there are two possible values ( and ).
Of course, as said by HallsofIvy, you could find the roots easier by solving for x in the given equation. Anyways, there isn't much more than that.
(Note: it's not by coincidence that the x position of the vertex turned out to be 2. The function given is written in a special form, which I can't recall its name, where the number substracting the x and the independent term are the vertex coordinates x and y repectively)
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